Quick but detailed way to see that $i^\#:\mathcal{O}(\mathbb{A}^2)\to \mathcal{O}(U)$ is an isomorphism? Consider the affine scheme $\mathbb{A}^2=(Spec \mathbb{C}[x,y],\mathcal{O})$ and the open subscheme $U=(U,\mathcal{O}|_U)$ where $U=Spec \mathbb{C}[x,y]\setminus\{(x,y)\}$. We have a natural inclusion $i:U\hookrightarrow \mathbb{A}^2$ where $i:U\to Spec\mathbb{C}[x,y]$ is just the inclusion and for all open $V\subset Spec\mathbb{C}[x,y]$ we have the map
$$i^\#_V:\mathcal{O}(V)\to (i^*\mathcal{O}\mid_U)(V)=\mathcal{O}\mid_U(i^{-1}(V))=\mathcal{O}(V\cap U)$$
given by the restriction $res_{V,V\cap U}:\mathcal{O}(V)\to \mathcal{O}(V\cap U)$. I want to determine $i^\#(Spec\mathbb{C}[x,y])$, i.e. the map
$$res_{Spec\mathbb{C}[x,y],U}:\mathcal{O}(Spec\mathbb{C}[x,y])\to \mathcal{O}(U)$$
Here is my approach: I know that $\mathcal{O}(U)\cong\mathbb{C}[x,y]$. This isomorphism is realised by noting that by definition of a sheaf $\mathcal{O}(U)$ can be identified with pairs of sections $(a,b)\in \mathcal{O}(D(x))\times \mathcal{O}(D(y))$ such that $a|_{D(xy)}=b|_{D(xy)}$ and this given an isomorphism $\mathcal{O}(U)\cong \mathbb{C}[x,y]\subset \mathcal{O}(D(xy))=\mathbb{C}[x,y,1/x,1/y]$. 
In other words, the image of an element $s\in\mathcal{O}(U)$ under this isomorphism $\mathcal{O}(U)\cong\mathbb{C}[x,y]$ is $s|_{D(xy)}$.
So to see that $res_{Spec\mathbb{C}[x,y],U}$ is an isomorphism, we note that $x|_{D(xy)}=x\in \mathbb{C}[x,y]\subset \mathcal{O}(D(xy))$ and similarly for $y$, which proves the claim.
Now I feel like I'm doing way to much work for this. I would appreciate a quicker way of seeing this, that is still precise and uses just the definitions of an affine scheme (so no argument like "a polynomial function on $\mathbb{C}^2\setminus 0$ can be extend to a polynomial function on $\mathbb{C}^2$ or something like that).
 A: I don't know if this is what you are looking for, but this presentation made the proof clearer for me. In particular, it gives a good justification as to why the isomorphism $\mathcal{O}(U) \simeq k[x,y]$ you wrote down is actually the restriction map.
Let $U = \mathbf{A}^2 \smallsetminus \{0\}$, which we can cover by $V_1 = \mathbf{A}^2 \smallsetminus \{x=0\}$ and $V_2 = \mathbf{A}^2 \smallsetminus \{x=0\}$. Let $V_{12} = V_1 \cap V_2$ be their intersection. By the sheaf property, the sequence
$$
  0 \longrightarrow \mathcal{O}(U) \longrightarrow \mathcal{O}(V_1) \times \mathcal{O}(V_2) \longrightarrow \mathcal{O}(V_{12})
$$
is an exact sequence (of abelian groups), where the maps are respectively given by
$$
  f \longmapsto (f\rvert_{V_1},f\rvert_{V_2}) \quad \text{and} \quad (g_1,g_2) \mapsto g_1\rvert_{V_{12}} - g_2\rvert_{V_{12}}.
$$
Since $\mathcal{O}(V_1) = k[x^{\pm1},y]$, $\mathcal{O}(V_2) = k[x,y^{\pm1}]$, and $\mathcal{O}(V_{12}) = k[x^{\pm1},y^{\pm1}]$, we have the commuative diagram
$$\require{AMScd}\begin{CD}
  0 @>>> k[x,y] @= k[x,y] @>>> 0\\
  @. @VV\rho_{\mathbf{A}^2,U}V @VV(\rho_{\mathbf{A}^2,V_1},\rho_{\mathbf{A}^2,V_2})V @VVV\\
  0 @>>> \mathcal{O}(U) @>>> k[x^{\pm1},y] \times k[x,y^{\pm1}] @>\varphi>> k[x^{\pm1},y^{\pm1}]
\end{CD}$$
where $\rho_{W,Z}$ denotes the restriction of a section from an open set $W$ to an open set $Z$. To show that $\rho_{\mathbf{A}^2,U}$ is an isomorphism, we use the snake lemma. Injectivity follows by injectivity of $(\rho_{\mathbf{A}^2,V_1},\rho_{\mathbf{A}^2,V_2})$. Surjectivity follows by noting that the map $\varphi$ induces an injection of the cokernel of $(\rho_{\mathbf{A}^2,V_1},\rho_{\mathbf{A}^2,V_2})$ into $k[x^{\pm1},y^{\pm1}]$, which follows by the exactness of
$$
  0 \longrightarrow k[x,y] \longrightarrow k[x^{\pm1},y] \times k[x,y^{\pm1}] \longrightarrow k[x^{\pm1},y^{\pm1}],
$$
which you already know. $\blacksquare$
